Mathematics Question and Solution
Let BC be a (square side).
tanb = 1/2
b = atan(½)°
sinatan(½) = c/a
c = 0.4472135955a
d = 2c
d = 0.894427191a
d is CE.
((½)a)² = 2²+e²
e² = (a²/4)-4
e = √(¼(a²-16))
e = ½√(a²-16)
f = e+(a/2)
f = ½√(a²-16)+(a/2)
f = ½(√(a²-16)+a)
f is CF.
It implies;
Observing Pythagoras Rule.
d² = 2²+f²
(0.894427191a)² = 4+(½(√(a²-16)+a))²
0.8a² = 4+¼((a²-16)+2a√(a²-16)+a²)
0.8a² = 4+¼(a²-16)+½(a√(a²-16))+¼(a²)
0.8a²-½(a²) = ½(a√(a²-16))
1.6a²-a² = a√(a²-16)
0.6a² = a√(a²-16)
0.6a = √(a²-16)
0.36a² = a²-16
16 = a²-0.36a²
16 = 0.64a²
1600 = 64a²
200 = 8a²
25 = a²
It implies;
Area ABCD (Area of the square), a² is;
a² = 25 square units.
Where a, the square side is 5 units.
